Why does $\left\{ \left( \frac{1}{n},\frac{1}{m} \right) : n,m \in Z^+ \right\}$ have Jordan measure $0$? 
Let $$A= \left\{ \left( \frac{1}{n},\frac{1}{m} \right) : n,m \in Z^+ \right\}$$

$Z^+$ denotes positive integers. How come this set has a zero area? 
Interior is definitely zero area since it doesn't have any interior points. How can I now prove that boundary has zero area? There exists irrationals in the boundary, that's why. How come irrationals of the boundary doesn't form some area?
 A: First: let $K= \{ 1/n : n \in \Bbb{Z}^+\}$. Obviously $A \subset K \times (0,1]$, so it is enough to prove that $K \times (0,1]$ has zero area. This is equivalent to say that $\forall \varepsilon > 0$ there exist finitely many rectangles $R_i$ covering $K \times (0,1]$ and whose total area is smaller than $\varepsilon$.
Now, fix any $0< \varepsilon <1$, and consider $R_0 = [0, \varepsilon^2] \times [0,1]$ whose area is $\varepsilon^2$. For all integers $1 \le n < \varepsilon^{-2}$ (there are finitely many: precisely $[\varepsilon^{-2}]$ many) consider the rectangle $$R_n=[1/n - \varepsilon^3 , 1/n + \varepsilon^3] \times [0,1]$$
whose area is $2 \varepsilon^3$. The total area of these rectangles is
$$ \varepsilon^2+ [\varepsilon^{-2}]2\varepsilon^3 < \varepsilon^2+ \varepsilon^{-2}2\varepsilon^3 = 2 \varepsilon + \varepsilon^2$$
which is arbirarily small.
If you note that $A \subset K \times (0,1] \subset \bigcup_{i=0}^{[\varepsilon^{-2}]} R_i$, then it is clear that they have zero Jordan measure.
A: Consider the set
$$A:=\left\{\left({1\over m_1},{1\over m_2},\ldots,{1\over m_d} \right)\>\biggm|\>m_i\in{\mathbb N}_{\geq1} \ (1\leq i\leq d)\right\}\subset{\mathbb R}^d\ .$$
Claim. The Jordan content, or $d$-dimensional Jordan measure, of this set is zero.
Proof. Let an $\epsilon>0$ be given. Consider the $d$ plates
$$Q_i:=\left\{x\in[0,1]^d\>\biggm|\>0\leq x_i\leq {\epsilon\over 2d}\right\}$$
of volume ${\epsilon\over 2d}$ each, and put $N:=\bigl\lfloor{2d\over\epsilon}\bigr\rfloor$. If $x:=\left({1\over m_1},{1\over m_2},\ldots,{1\over m_d} \right)\in A$ but $x\notin\bigcup_{i=1}^d Q_i$ then  necessarily $m_i\leq N$ for $1\leq i\leq d$. It follows that there are $N^d$ such bad points $x$. Cover each of them with a cube $C_k$ of sidelength $\epsilon'>0$ so small that the total volume of these cubes is $<{\epsilon\over2}$. It follows that
$$A\subset\bigcup_{i=1}^d Q_i\ \cup\ \bigcup_{k=1}^{N^d} C_k\ ,$$
whereby the finitely many boxes on the right hand side have total volume $<\epsilon$.
